Rip

$\operatorname{Hom}_R(M,N)$ is not an $R$-module if $M$ and $N$ are

the annihilator of $R/I$ is not necessarily $I$

@B.Mehta I don’t think you’d actually find the result in here, but given the author I felt obliged to link this: elsevier.com/books/random-matrices/lal-mehta/978-0-12-088409-4

@Semiclassical Haha :D

Even if $I+J=(1)$ for two-sided ideals, we don't necessarily have $I \cap J = IJ$

sums of nilpotent elements need not be nilpotent. In particular, the set of all nilpotent elements is not an ideal

over a division ring, polynomials in one variable can have infinitely many roots

When A is a ring, then and $a \in A$, then the evaluation map $A[x] \to A, f \mapsto f(a)$ doesn't have to be a ring homomorphism (it works iff $a$ is central, basically because $x$ commutes with every element of $A$ inside $A[x]$)

Oh that's gonna mess me up hard

even if you have a domain, there might not be a canonical way to embed it into a division ring

some LA carries over to division rings, but determinants don't work

@MatheinBoulomenos The last one about the evaluation map rekt me hard in my algebra test a week ago

@Perturbative interesting, what kind of algebra course are you taking?

The question was to give an example of a polynomial ring $R[x]$ over a base ring $R$ and polynomials $h(x), f(x), g(x)$ such that $h(x) = f(x)g(x)$ but $h(a) \neq f(a)g(a)$ for some $a \in R$

What you said makes sense because $R[x]$ is commutative (always?) but $R$ need not be

$R[x]$ is certainly not commutative if $R$ isn't

but $x$ is a central element

@MatheinBoulomenos Just a basic intro course, groups, rings and fields up until Kronecker's theorem

@MatheinBoulomenos Oh I see

I'd take $f$ constant with some value that doesn't commute with $a$ and $g(x)=x$

Yeah the solution we got given afterwards is basically what you just described

technically, one also needs to specify that we evaluate with coefficients on the left since that's not the same as evaluating with coeffients on the right

@Daminark do you want to see some weird noncommutative ring?

Sure!

Let $k$ be a field and $V$ be an infinite dimensional vector space over $k$. Note that we have $V \cong V \oplus V$, fix such an isomorphism $f: V \to V \oplus V$ and denote the components of $f$ by $f_1$ and $f_2$. Let $R=\operatorname{End}_R(V)$ (basically infinite matrices)

If we have $\varphi \in R$, we can compose with $f_1$ and $f_2$ from the right and this gives us again $k$-linear maps $V \to V$, i.e. elements of $R$. So we get a map $R \to R^2$ given by $\varphi \mapsto (\varphi \circ f_1, \varphi \circ f_2)$

With you so far

ah no wait

Okay so we can also define a map $R^2 \to R$ where we send $(\varphi_1,\varphi_2)$ to $v \mapsto (\varphi_1(f^{-1}(v))+\varphi_2(f^{-1}(v)))$

that's an inverse to the map above

and both these maps are acutally left $R$-linear

so $R\cong R^2$ as $R$-modules

and by induction $R^n \cong R^k$ for all $n,k \in \Bbb N$

Did you mean $f_i^{-1}$? Since $f^{-1}$ takes something from $V\oplus V$

ah yeah there was something wrong in that

@MatheinBoulomenos thanks for your help yesterday man! Saved my ass today lol

Also side note, the functional analysis book that I recommended to GFauxPas earlier is one you might like (since you mentioned you didn't like your functional class doing more numerical stuff). Functional Analysis, Spectral Theory, and Applications

Our professor probably would've used it for our class if he knew about it then

$\varphi_1, \varphi_2$, we take $v$ then $f(v) = (v_1,v_2)$ and we map that to $\varphi_1(v_1)+\varphi_2(v_2)$

not sure how to write that down in one expression

Hmm, $\phi_1(f_1(v)) + \phi_2(f_2(v))$?

Err, varphi I guess, I'm always too lazy to write var down :P

hmm, something doesn't seem right still

it's easier if you think in Hom-Sets. $R = Hom_k(V,V) \cong Hom_k(V,V \oplus V) \cong Hom_k(V,V) \oplus Hom_k(V,V) = R^2$ and since we only changed things in the second component, if we compose (i.e. multiply) we can pull that out, so it's left $R$-linear

I wanted to write down the maps explicitly but I messed up somehow

@Daminark yes

so this shows that "rank of a free module" doesn't need to make a whole lot of sense for general noncommutative rings

@ÍgjøgnumMeg no problem

Ah I see now

these computations canbe tricky

@Daminark so using a R metric you can dualize an o.n. basis to get an o.n. dual basis

and then you wedge those bois to get the volume form

easy exercize to show that it's actually a volume form

Ah I understand now why I messed up

If we compose in $Hom_k(V,V)$ in the left component, that's multiplication from the right

@Daminark then when $G$ is compact you can normalize the volume form and this actually gives you the legit Haar measure

hi peeps

hello old lady

hi peep

I see. Well, I don't quite know how Riemannian metrics work but I'm willing to take that

@Daminark I can tell you if you want to pay attention

on Lie groups they're particularly easy to understand

Aight, let's see the Lie groups case

Hopefully that'll give a decent framework

@Daminark So in general a Riemannian metric is an assignment of inner product to each tangent space in a "smooth" manner

Smooth in the sense that if $X,Y$ are smooth vector fields, then $g(X,Y)$ is a smooth function (this definition is obviously local)

So the way to get lots of metrics on Lie groups is to push forward an inner product on $T_eG=\mathfrak g$ via left translation

For $a\in G$ we have the smooth left translation map $L_ag=ag$

hi professor

hi skull

Welcome back Ted!

Now, given a Riemannian metric $h$ on $G$, we say that $h$ is left invariant if $$h_b(x,y)=h_{ab}(d(L_a)x,d(L_a)y)$$

thanks, Demonark

how are you?

for all $a,b\in G, x,y\in T_bG$

$h_b$ meaning the inner product at the point $b\in G$?

yeah

Okay